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 Minimal area. (Posted on 2010-03-29)
Take some point V and draw two rays from it. Choose some other point W in between those two rays. Then, construct a line that touches both rays and passes through W.

Now, this line forms a closed triangle together with the two rays. The point W divides this line into two segments (x1, x2). What is the ratio of these two segments such that the area of the enclosed triangle is minimal?

Does this minimal area even exist?

 See The Solution Submitted by Vee-Liem Veefessional Rating: 4.6667 (3 votes)

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The ratio of x1 and x2 is wholly dependent on the positioning of W within the arc.  The area of the triangle is maximized as either adjacent angle approaches zero.  In theory, as the angle approaches zero, the adjacent side approaches parallelness with the moveable line segment containing point W.  And the area between two parallel lines in a plane is infinite.

To minimize the area, both angles must be the same.  Thus, an isosceles triangle will have the minimal area.  However, that status is independent of the position of W along that line segment.  Thus the ratio of x1 and x2 is solely dependent of the position of W on that line, but independent of the angles needed to minimize triangle area.

 Posted by hoodat on 2010-03-29 15:45:08

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